Atoms & Nuclei

Rutherford's $\alpha$ scattering experiment (1911): Alpha particles fired at a thin gold foil showed that most passed through undeflected, a few were scattered at large angles, and very rarely some bounced back (180 deg). Conclusions: (1) the atom is mostly empty space, (2) all positive charge and nearly all mass are concentrated in a tiny nucleus ($10^{-15}$ m), (3) electrons orbit the nucleus at relatively large distances (~$10^{-10}$ m).
The distance of closest approach for a head-on collision: d = $\frac{2kZe^{2}}{\text{KE}_{alpha}}$, where k = 9 x $10^{9}$ N $\frac{m^{2}}{C^{2}}$, Z = atomic number of target, e = 1.6 x $10^{-19}$ C, $\text{KE}_{alpha}$ = kinetic energy of $\alpha$ particle; [d] = [L], SI unit: m. This gives an upper limit on nuclear size.

Bohr model postulates: (1) Electrons revolve in fixed circular orbits (stationary states) without radiating energy.
(2) Angular momentum is quantized: L = nh/($2\pi$) = n hbar, where n = 1, 2, 3, ...
(3) Photons are emitted or absorbed during transitions: $E_{photon}$ = h $\nu$ = $E_{i}$ - $E_{f}$.

Bohr model results for hydrogen-like atoms (atomic number Z):

• Radius: $r_{n}$ = 0.529 x $n^{2}$/Z angstrom = $a_{0} \; n^{2}$/Z, where $a_{0}$ = 0.529 angstrom (Bohr radius); [r] = [L], SI unit: m. Radius increases as $n^{2}$ and decreases with Z.
• Velocity: $v_{n}$ = 2.18 x $10^{6}$ x Z/n m/s; [v] = [L $T^{-1}$], SI unit: m/s. Velocity decreases with n.
• Energy: $E_{n}$ = -13.6 $\frac{Z^{2}}{n^{2}}$ eV; [E] = [M $L^{2} \; T^{-2}$], SI unit: J or eV. The negative sign indicates a bound state. Ground state (n = 1) is most tightly bound.

Energy relationships: KE = -$E_{n}$ = 13.6 $\frac{Z^{2}}{n^{2}}$ eV (positive).
PE = $2E_{n}$ = -27.2 $\frac{Z^{2}}{n^{2}}$ eV (negative).
Total energy E = KE + PE.
Key relation: KE = -E (magnitude of total energy), PE = 2E (twice the total energy, negative).

Hydrogen spectrum: 1/$\lambda$ = $RZ^{2}$(1/$n_{1}^{2}$ - 1/$n_{2}^{2}$), where R = 1.097 x $10^{7} \; m^{-1}$ is the Rydberg constant; [R] = [$L^{-1}$]. Spectral series:

• Lyman ($n_{1}$ = 1, $n_{2}$ = 2, 3, ...): UV region.
  Shortest $\lambda$ (series limit): $n_{2}$ = infinity.
• Balmer ($n_{1}$ = 2, $n_{2}$ = 3, 4, ...): Visible region (most tested in NEET).
• Paschen ($n_{1}$ = 3, $n_{2}$ = 4, 5, ...): Infrared region.
• Brackett ($n_{1}$ = 4) and Pfund ($n_{1}$ = 5): Far infrared.

The number of spectral lines possible from the nth energy level: N = n(n - 1)/2.

Nucleus: Composed of Z protons and (A - Z) neutrons, where A = mass number, Z = atomic number.
Nuclear radius: R = $R_{0}$ A^($\frac{1}{3}$), where $R_{0}$ = 1.2 fm = 1.2 x $10^{-15}$ m; [R] = [L]. Nuclear density is constant (~2.3 x $10^{17}$ kg/$m^{3}$) for all nuclei because volume proportional to A and mass proportional to A.

Mass defect: $\Delta$ m = [Z $m_{p}$ + (A - Z) $m_{n}$] - M, where M is the actual nuclear mass; [$\Delta$ m] = [M], SI unit: kg or u (1 u = 931.5 MeV/$c^{2}$).
Binding energy: BE = $\Delta$ m x 931.5 MeV; [BE] = [M $L^{2} \; T^{-2}$], SI unit: MeV or J. The BE per nucleon curve peaks at Fe-56 (~8.75 MeV/nucleon): lighter nuclei can release energy by fusion (combining), heavier nuclei by fission (splitting).

Radioactive decay: N = $N_{0}$ e^(-$\lambda$ t), where $\lambda$ is the decay constant; [$\lambda$] = [$T^{-1}$], SI unit: $s^{-1}$.
Half-life: $\frac{t_{1}}{2}$ = 0.693/$\lambda$ = ln 2/$\lambda$; [$\frac{t_{1}}{2}$] = [T], SI unit: s.
After n half-lives: N = $\frac{N_{0}}{2}$^n.
Activity: A = $\lambda$ N = $A_{0}$ e^(-$\lambda$ t); [A] = [$T^{-1}$], SI unit: becquerel (Bq).
Mean life: $\tau$ = 1/$\lambda$ = $\frac{t_{1}}{2}$/0.693.

Decay types: Alpha (emits He-4: A decreases by 4, Z by 2), Beta-minus (neutron to proton + electron + antineutrino: Z increases by 1, A unchanged), Gamma (nucleus de-excites by emitting photon: no change in A or Z).

The key testable concept is the Bohr model energy level calculations, spectral series identification, and half-life calculations using the radioactive decay law.

Solved Numericals

N₁. Find the radius, velocity, and energy of an electron in the 3rd orbit of hydrogen. Compare with ground state values.

Given: Z = 1 (hydrogen), n = 3, $a_{0}$ = 0.529 angstrom, $v_{1}$ = 2.18 x $10^{6}$ m/s, $E_{1}$ = -13.6 eV.

Radius: $r_{3}$ = $a_{0}$ x $n^{2}$/Z = 0.529 x $\frac{9}{1}$ = 4.761 angstrom = 4.761 x $10^{-10}$ m. Compared to ground state: $r_{1}$ = 0.529 angstrom. Ratio $\frac{r_{3}}{r_{1}}$ = 9 (increases as $n^{2}$).

Velocity: $v_{3}$ = $v_{1}$ x Z/n = 2.18 x $10^{6}$ x $\frac{1}{3}$ = 7.27 x $10^{5}$ m/s. Compared: $v_{1}$ = 2.18 x $10^{6}$ m/s.
Ratio $\frac{v_{3}}{v_{1}}$ = $\frac{1}{3}$ (decreases as 1/n).

Energy: $E_{3}$ = $E_{1}$ x $\frac{Z^{2}}{n^{2}}$ = -13.6 x $\frac{1}{9}$ = -1.511 eV. Compared: $E_{1}$ = -13.6 eV. Ratio $\frac{E_{3}}{E_{1}}$ = $\frac{1}{9}$ (less tightly bound in higher orbits).

Also: $\text{KE}_{3}$ = -$E_{3}$ = +1.511 eV.
$\text{PE}_{3}$ = $2E_{3}$ = -3.022 eV. Check: $\text{KE}_{3}$ + $\text{PE}_{3}$ = 1.511 + (-3.022) = -1.511 eV = $E_{3}$. Verified.

N₂. Calculate the wavelength of the first line (longest wavelength) and series limit (shortest wavelength) of the Balmer series for hydrogen.

Balmer series: $n_{1}$ = 2. R = 1.097 x $10^{7} \; m^{-1}$.

First line (longest wavelength): $n_{2}$ = 3 (smallest energy transition in the series). 1/$\lambda$ = R(1/$2^{2}$ - 1/$3^{2}$) = 1.097 x $10^{7}$ ($\frac{1}{4}$ - $\frac{1}{9}$) = 1.097 x $10^{7}$ x $\frac{5}{36}$ = 1.097 x $10^{7}$ x 0.1389 = 1.524 x $10^{6} \; m^{-1}$. $\lambda$ = $\frac{1}{1}$.524 x $10^{6}$ = 6.563 x $10^{-7}$ m = 656.3 nm (red light — H-$\alpha$ line).

Series limit (shortest wavelength): $n_{2}$ = infinity. 1/$\lambda$ = R(1/$2^{2}$ - 0) = R/4 = 1.097 x $\frac{10^{7}}{4}$ = 2.7425 x $10^{6} \; m^{-1}$. $\lambda$ = $\frac{1}{2}$.7425 x $10^{6}$ = 3.646 x $10^{-7}$ m = 364.6 nm (near UV — series limit).

The Balmer series spans from 656.3 nm (red) to 364.6 nm (near UV). The visible lines are H-$\alpha$ (656.3 nm, red), H-$\beta$ (486.1 nm, blue-green), H-$\gamma$ (434.0 nm, violet), and H-$\delta$ (410.2 nm, deep violet).

N₃. A radioactive sample has a half-life of 10 days. Find: (a) fraction remaining after 30 days, (b) time for activity to drop to $\frac{1}{16}$ of initial, (c) decay constant and mean life.

Given: $\frac{t_{1}}{2}$ = 10 days.

(a) Number of half-lives in 30 days: n = t/$\frac{t_{1}}{2}$ = $\frac{30}{10}$ = 3. Fraction remaining: N/$N_{0}$ = $\frac{1}{2}$^n = 1/$2^{3}$ = $\frac{1}{8}$. So $\frac{1}{8}$ of the original sample remains (or $\frac{7}{8}$ has decayed).

(b) Activity drops to $\frac{1}{16}$ of initial: A/$A_{0}$ = $\frac{1}{16}$ = 1/$2^{4}$. Number of half-lives: n = 4. Time: t = n x $\frac{t_{1}}{2}$ = 4 x 10 = 40 days.

(c) Decay constant: $\lambda$ = 0.693/$\frac{t_{1}}{2}$ = 0.$\frac{693}{10}$ days = 0.0693 per day. In SI: $\lambda$ = 0.0693/(86400 s) = 8.02 x $10^{-7} \; s^{-1}$.

Mean life: $\tau$ = 1/$\lambda$ = $\frac{t_{1}}{2}$/0.693 = $\frac{10}{0}$.693 = 14.43 days.

Note: Mean life is always greater than half-life (by a factor of $\frac{1}{0}$.693 = 1.443).